Find the horizontal and vertical component of water pressure acting on the face of a tainter gate of 90° sector of radius 4 m as shown in the figure. Take width of gate as unity (1 m).

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Tainter Gate Fluid Pressure Problem

Problem Statement

Find the horizontal and vertical component of water pressure acting on the face of a tainter gate of 90° sector of radius 4 m as shown in the figure. Take width of gate as unity (1 m).

Given Data

  • Radius of Gate, \( R = 4 \, \text{m} \)
  • Sector Angle, \( = 90^\circ \)
  • Width of Gate, \( b = 1 \, \text{m} \) (unity)
  • Density of Water, \( \rho = 1000 \, \text{kg/m}^3 \)
  • Acceleration due to Gravity, \( g = 9.81 \, \text{m/s}^2 \)

Diagram of Tainter Gate

The 90° tainter gate submerged in water.

Diagram of the Tainter Gate

Solution

To find the resultant force on the curved gate, we calculate its horizontal and vertical components separately.

1. Horizontal Force (\(F_x\))

The horizontal force is the total pressure on the projected area of the curved surface onto a vertical plane (area ADB).

$$ \text{Projected Height (AB)} = 2 \times AD = 2 \times (R \sin 45^\circ) $$ $$ \text{Projected Height (AB)} = 2 \times (4 \times 0.707) $$ $$ = 5.656 \, \text{m} $$
$$ A_{proj} = \text{Projected Height} \times \text{Width} $$ $$ = 5.656 \, \text{m} \times 1 \, \text{m} $$ $$ = 5.656 \, \text{m}^2 $$

The depth of the centroid of this projected area from the free surface is:

$$ \bar{h} = \frac{\text{Projected Height}}{2} $$ $$ = \frac{5.656}{2} $$ $$ = 2.828 \, \text{m} $$

The horizontal force is calculated as:

$$ F_x = \rho g A_{proj} \bar{h} $$ $$ F_x = 1000 \times 9.81 \times 5.656 \times 2.828 $$ $$ F_x = 156911 \, \text{N} $$

2. Vertical Force (\(F_y\))

The vertical force is equal to the weight of the water enclosed by the curved surface (portion ACBDA).

The area of this portion is the area of the sector ACBOA minus the area of the triangle ABO.

$$ \text{Area of Sector} = \frac{90^\circ}{360^\circ} \times \pi R^2 $$ $$ = \frac{\pi (4)^2}{4} = 4\pi \, \text{m}^2 $$
$$ \text{Area of } \triangle ABO = \frac{1}{2} \times \text{base} \times \text{height} $$ $$ = \frac{1}{2} \times 4 \times 4 = 8 \, \text{m}^2 $$
$$ \text{Area (ACBDA)} = \text{Area of Sector} - \text{Area of } \triangle ABO $$ $$ = (4\pi - 8) \, \text{m}^2 $$ $$ \approx 4.566 \, \text{m}^2$$

The vertical force is the weight of the water in this volume:

$$ F_y = \rho g \times (\text{Area}) \times \text{Width} $$ $$ F_y = 1000 \times 9.81 \times (4\pi - 8) \times 1 $$ $$ F_y \approx 44796 \, \text{N} $$
Final Result:

The horizontal component of the force is \( F_x \approx 156911 \, \text{N} \) or \( 156.91 \, \text{kN} \).

The vertical component of the force is \( F_y \approx 44796 \, \text{N} \) or \( 44.80 \, \text{kN} \).

Explanation of Concepts

Horizontal Force on Curved Surface: The horizontal component of the hydrostatic force on any curved surface is always equal to the force on the vertical projection of that surface. This allows us to simplify the problem by analyzing a simple flat plate.

Vertical Force on Curved Surface: The vertical component is equal to the weight of the fluid volume directly above (or enclosed by) the curved surface, extending up to the free surface. This force acts through the centroid (center of gravity) of that fluid volume.

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